Question

ii) u2 + 9v2 + 2w2 – 6uv - 6√2vw + 2√2wu​

Answer 1

Answer:

Solution. It is obvious from the equations of the lines bounding R that we should use the transformation

x − y = u and x + y = v. If we add these two equations we get 2x = u + v or x = (u + v)/2. If we subtract

the first from the second we get 2y = v − u or y = (v − u)/2. The equations of the lines also indicate that the

bounds on u are 0 ≤ u ≤ 2 and the bounds on v are 0 ≤ v ≤ 3. Next find the Jacobian.

∂(x, y)

∂(u, v)

=

¯

¯

¯

¯

1

2

1

2

−

1

2

1

2

¯

¯

¯

¯ =

1

4

+

1

4

=

1

2

Now we can evaluate the integral. Note that x

2 − y

2 = (x − y)(x + y) = uv.

Z Z

R

(x + y)e

x

2−y

2

dA =

Z 3

0

Z 2

0

veuv

·

1

2

du dv

=

1

2

Z 3

0

e

uv

¯

¯

¯

¯

2

0

dv

=

1

2

Z 3

0

(e

2v − 1) dv

=

1

2

µ

1

2

e

2v − v

¶ ¯

¯

¯

¯

3

0

=

1

2

·µ1

2

e

6 − 3

¶

−

1

2

¸

=

1

4

e

6 −

7

4